Prove that if $a$ is a root of $fg$, $a$ is a root of $f$ or $a$ is a root of $g$.

Question

Let $R$ be an integral domain, $f, g \in R[x], a \in R$. Prove that if $a$ is a root of $fg$, then $a$ is a root of $f$ or $a$ is a root of $g$.

My attempt (edited):

pf. Let $R$ be an integral domain, $f, g \in R[x], a \in R$. If $a$ is a root of $fg$, then $fg(a)$ = 0. Since $R$ is an integral domain, a nonzero non-unit $r \in R$ is irreducible if for all $a, b \in R$ such that $r = ab$, either $a$ or $b$ is a unit. Hence if $a$ is a root of $fg$, $b$ must be a unit. Thus, $b^-1 \in R$ and so $b^-1fg = abb^-1$. Hence, $a = b^-1fg$. Thus, $fg \vert a$ so $a$ must divide $f$ or $g$.

I am not too sure about the last couple of statements. Would appreciate any feedback!

Answer 1

If $a \in R$ is a root of

$f(x)g(x) \in R[x], \tag 1$

then

$f(a)g(a) = fg(a) = 0; \tag 2$

now if

$f(a) \ne 0 \ne g(a), \tag 3$

then since $R$ is an integral domain we have

$f(a)g(a) \ne 0; \tag 4$

but this contradicts (2); thus (3) is false and hence

$f(a) = 0 \; \text{or} \; g(a) = 0; \tag 5$

that is, $a$ is a root of at least one of $f(x)$, $g(x)$.

Answer 2

Let $R$ be an integral domain, $f,g\in R[x],$ $a\in R.$ If $a$ is a root of $fg,$ then $fg(a) = 0.$ Since $R$ is an integral domain, a nonzero non-unit $r\in R$ is irreducible if for all $a,b\in R$ such that $r=ab,$ either $a$ or $b$ is a unit.

All of this is fine.

Hence if $a$ is a root of $fg$, $b$ must be a unit.

A stylistic comment: at this point, you haven't defined what $b$ is, but you start talking about it as if your reader will know what it is. How does $b$ relate to $f,$ $g,$ and $a$? You defined what it means for an element $r\in R$ to be irreducible in general, but now you seem to be applying it without explaining what you're assuming is irreducible.

Thus, $b^{−1}\in R$ and so $b^{-1}fg = abb^{-1}$. Hence, $a=b^{-1}fg$.

Where did the equation $b^{-1}fg = abb^{-1} = a$ come from? This is equivalent to saying that $fg = ab,$ which was not an assumption at any point of the problem. Remember that at the beginning of the problem, we assume that $fg(a) = f(a)g(a) = 0$ (remember also that if $p\in R[x]$ is a polynomial and $r\in R,$ then $p(r)$ does not mean $p\cdot r$ (the product of the polynomial $p$ and the constant polynomial $r$) but rather the evaluation of $p$ at $r$ (what you get when you replace all the $x$'s in $p(x)$ with $r$'s).

Thus, $fg\mid a$ so $a$ must divide $f$ or $g$.

Indeed, $a = b^{-1}fg$ implies that $fg$ divides $a.$ However, to conclude from this that $a$ divides $f$ or $g$ requires that $a$ be a prime element, which again is not an assumption of the problem. Moreover, even if you could conclude that $a$ divides $f$ or $g,$ this does not show that $a$ is a root of $f$ or of $g,$ because the definition of $a\in R$ being a root of $f$ is not that $a$ divides $f,$ but rather that $f(a) = 0.$

In order to actually prove the statement, you should proceed as described in Robert Lewis' answer. Either try a proof by contradiction assuming that $a$ is not a root of $f$ or $g,$ so that $f(a)\neq 0$ and $g(a)\neq 0$ and then use the fact that $R$ is an integral domain to conclude that $fg(a)\neq 0,$ so that $a$ is not a root of $fg$ (contradiction!).